Re: Set theory theorem: f_* = g_* iff f^* = g^* iff f = g.
From: William Elliot (marsh_at_privacy.net)
Date: 07/04/04
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Date: Sun, 4 Jul 2004 06:29:44 -0700
On Sun, 4 Jul 2004, Adam wrote:
> If f is a map, then f^* is the inverse image, and f_* is the image.
>
I use regular notation, f^-1(A) for f^*(A) and f(A) for f_*(A).
> Theorem. Let f, g: A -> B be maps. Think of these maps as inducing maps f_*,
> g_*: P(A) -> P(B), and maps f^*, g^*: P(B) -> P(A). Then f_* = g_* iff f^* =
> g^* iff f = g.
>
> Proof. We will show that f = g implies f^* = g^*, that f^* = g^* implies f_*
> = g_*, and finally that f_* = g_* implies f = g. The domains and codomains
> of the maps are respectively the same, and so we only need to show the
> mapping of elements. We will let X in P(A), and so X <= A for all the
> sections below.
>
> We begin by assuming f = g, and then show that f^* = g^*. Suppose x in
> f^*(X); we will show x in g^*(X); Then f(x) in X. Since f = g, g(x) in X,
> and thus x in g^*(X), as desired.
> Now suppose x in g^*(X); we will show x in f^*(X). Then g(x) in X. Since
> f = g, f(x) in X, and thus x in f^*(X), as desired. Therefore, f = g implies
> f^* = g^*.
>
f = g ==> f_* = g_* and f^-1 = g^-1 is immediate by definition of equality
and that f_* and f^-1 are defined from f. To wit, assuming f = g
f(U) = { f(x) | x in U } = { g(x) | x in U } = g(U)
f^-1(U) = { x | f(x) in U } = { x | g(x) in U } = g^-1(U)
> In this part we assume f^* = g^*, and then show this implies f_* = f_*.
> Suppose x in f_*(X); we will show x in g_*(X). Then x = f(p) for some p in
> X. Since p in X, f(p) in f_*(X), and so p in f^*(f_*(X). Since f^* = g^*, p
> in g^*(f_*(X)). Thus g(p) in f_*(X). Since p in X, g(p) in g_*(X), which
> implies p in g^*(g_*(X)). Thus x in f_*(g^*(g_*(X))), and so x in
> f_*(f^*(g_*(X))), by using the assumption again. Then x = f(w) for some w in
> f^*(g_*(X)). And so f(w) in g_*(X). Equating with x we see that x in g_*(X),
> as desired.
> We now follow a simliar process, typograghically replacing the "f"'s
> with "g"'s. Suppose x in g_*(X); we will show x in f_*(X). Then x = g(p) for
> some p in X. Since p in X, g(p) in g_*(X), and so p in g^*(g_*(X). Since f^*
> = g^*, p in f^*(g_*(X)). Thus f(p) in g_*(X). Since p in X, f(p) in f_*(X),
> which implies p in f^*(f_*(X)). Thus x in g_*(f^*(f_*(X))), and so x in
> g_*(g^*(f_*(X))), by using the assumption again. Then x = g(w) for some w in
> g^*(f_*(X)). And so g(w) in f_*(X). Equating with x we see that x in f_*(X),
> as desired. This result, combined with the previous paragraphs result, shows
> that f^* = g^* implies f_* = g_*.
>
No thanks, much too long. Here's adirect approach:
Assume f^-1 = g^-1. Now when x in A
f(x) in { f(x) }
x in f^-1({ f(x) }) = g^-1({ f(x) })
x in g^-1({ f(x) })
g(x) in { f(x) }; g(x) = f(x)
Thus for all x in A, f(x) = g(x); f = g
I leave for you to know
f = g ==> f_* = g_*
> The final part to prove is that f_* = g_* implies f = g, which we now
> do. Suppose x in A. Then f(x) in f_*({x}). We assume that f_* = g_*, and
> thus f(x) in g_*({x}). So f(x) = g(p) for some p in {x}. Since x is the only
> element, p = x, and so f(x) = g(x), as desired. Therefore, f_* = g_* implies
> f = g, and the proof is complete. QED.
>
If f_* = g_*, then
{ f(x) } = f({ x }) = g({ x }) = { g(x) }
thus for all x, f(x) = g(x) and f = g
Basically the same approach as yours.
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